Graph Decomposition and graph factor

If a graph G is H-decomposable does it imply that G has H-factor?

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No. An $H$-factor is a set of vertex-disjoint copies of $H$ that partition the vertex set of a graph, while an $H$-decomposition is a set of edge-disjoint copies of $H$ that partition the edge set of a graph.
For example, let $H = K_3$ and let $G$ be the $5$-vertex graph consisting of two triangles with a point in common. Then $G$ clearly has an $H$-decomposition, but, since $3 \nmid 5$, it cannot have an $H$-factor.
Say i have a graph G with 3m vertices, and that G is $C_3$-decomposable. Can i say now that G has $C_3_-factor? – kim_kibun Feb 21 '13 at 14:44 @kim_kibun Again, not necessarily. Let$G$be the$6$-vertex graph obtained by taking a triangle and adding three vertices, each adjacent to two of the vertices of the triangle. (More formally, let the vertices of the triangle be$u$,$v$, and$w$, and let the other three vertices be$a$,$b$, and$c$. Let$a$be adjacent to$u$and$v$,$b$to$v$and$w$, and$c$to$w$and$u$.) Then$G$has a$K_3$-decomposition but no$K_3\$-factor. – Andrew Uzzell Feb 21 '13 at 18:13